如何顯示MySQL中GROUP BY計數大於1的所有結果？

Question

這是我的MySQL查詢及其返回的內容：

SELECT email, COUNT(*) AS num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL    
GROUP BY email
ORDER BY dateadded DESC;

我只想返回num> 1的結果。我試圖像這樣更改查詢，但是說num不是可識別的列是行不通的：

SELECT email, COUNT(*) AS num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL
AND num > 1
GROUP BY email
ORDER BY dateadded DESC;

如果num> 1，如何返回結果？

Answer 1

在GROUP BY子句之后和ORDER BY子句之前，添加以下內容：

HAVING COUNT(*) > 1 

在准備好所有行之后，在LIMIT子句之前， HAVING子句幾乎在執行計划中最后一次應用。 這對於在訪問行時無法檢查的條件（但僅在訪問行之后才可以檢查的條件）最為有用，例如像COUNT（*）之類的聚合函數，盡管它可以用於非聚合。

Answer 2

使用HAVING子句。 或另一種方式：

SELECT * FROM
 (SELECT email, COUNT(*) AS num  
  FROM collectors_users
  WHERE subscribed != 'NO'
  AND lastLogin IS NULL    
  GROUP BY email
  ORDER BY dateadded DESC;
  ) T
WHERE num>1

Answer 3

我認為您應該使用HAVING

SELECT email, COUNT(*) AS num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL
GROUP BY email
ORDER BY dateadded DESC
HAVING COUNT(*) > 1; 

這個問題可以幫助你理解為什么HAVING而不是WHERE

在哪里與擁有

Answer 4

SELECT email, COUNT(*) as num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL
GROUP BY email
HAVING COUNT(*) > 1

Answer 5

您可以使用Haveing子句。

SELECT email, COUNT(*) AS num  
FROM collectors_users
WHERE subscribed != 'NO'
AND lastLogin IS NULL
GROUP BY email
HAVING count(*) > 1
ORDER BY dateadded DESC;

有關更多信息： http : //www.w3schools.com/sql/sql_having.asp